2.7 Radiation
2.7.1 Definitions and laws
Heat transfer by conduction and convection required the existence of a material
medium, either a solid or a fluid. However, it is not required in heat transfer by
radiation.
Radiation can travel through an empty space at the speed of light in the form of
an electromagnetic wave. As shown in the electromagnetic spectrum in Fig. 2.7.1,
thermal radiation covers the range of wavelength from 0.1~100 μm.
2.7.1.1 Absorptivity
Thermal radiation impinging on the surface of an opaque solid is either absorbed
or reflected. The absorptivity is defined as the fraction of the incident radiation that
45
is absorbed:
q(a)
  (i )
q
[2.7.1]
Where q(a) is the energy absorbed per unit area per unit time and q(i) is the
energy impinging per unit area per unit time.
Let us define ql(a) and ql(i) such that ql(a)dl and ql(i)dl represent respectively
the absorbed and incident energies per unit area per unit time in the wavelength
range l to l+dl. The monochromatic absorptivity l is defined as :
ql dl ql
 l  (i )  (i )
ql dl ql
(a)
(a)
[2.7.2]
For any real body αλ< 1 and depend on l. A graybody is a hypothetical
one for which αλ < 1 but independent of l and temperature. The limiting case
of αλ = 1 for all l. and temperature is known as a blackbody. In other words,
a blackbody absorbs all the incident radiation.
2.7.1.2 Emissivity
q (e)
The emissivity of a surface is defined as   ( e )
qb
[2.7.3]
Where q(e) and qb(e) are the energies emitted per unit area per unit time by a
real body and a blackbody, respectively. Table 2.7-1 lists the emissivities of some
Material. These averaged value have been used widely even though the emissivity
46
actually depends on the wavelength and the angle of emission.
Let us define ql(e) and qbl(e) such that ql(e)dl and qbl(e)dl represent respectively
the energies emitted per unit area per unit time in the wavelength range l to l+dl
of a real body and blackbody. The monochromatic emissivity l is defined as :
q dl
q
 l  l (e)  l (e)
qbl dl qbl
(e)
(e)
[2.7.4]
l=1 for blackbody and <1 for a real body.
47
2.7.1.3 Kirchhoff’s law
Consider the body enclosed in the cavity shown in Fig. 2.7.2.
Suppose the body is at the same temperature as the wall of the
cavity, that is, the two are at thermodynamic equilibrium with each
other. Since there is no net heat transfer from the cavity wall to The
body, the energy emitted by the body must be equal to the energy absorbed.
q ( e ) A  q (i ) A
[2.7.5]
Where q(e) is the energy emitted per unit area of the body per unit time, A the surface
area of the body, q(i) the energy impinging per unit area of the body per unit time, and
 the absorptivity of the body. If the body is replaced by a blackbody, Eq.[2.7.5]
becomes
[2.7.6]
(e)
(i )
qb
Aq A
q (e)
Dividing Eq. [2.7.5] by Eq.[2.7.6], we get

(e)
qb
[2.7.7]
Substituting Eq. [2.7.7] into Eq.[2.7.3], we have
[2.7.8]
 
48
This is Kirchhoff’s law, which states that for a system in thermodynamic equilibrium
the emissivity and absorptivity are the same. Following a similar procedure, we can
show that
 l  l
[2.7.9]
2.7.1.4 Plank’s distribution law
Plank derived the following equation for the energy emitted by a blackbody as a
function of the wavelength and temperature:
[2.7.10]
qb(le )
Where
2 c 2 hl 5

exp(ch / k lT )  1
T = temperature in K
c = speed of light
h = Plank’s constant
k = Boltzmann’s constant
49
2.7.1.5 Stefan-Boltzmann law
The Plank distribution law can be integrated over wavelengths from zero to infinity
to determine the total emissive energy of a blackbody:
(e)
b
q


0
2 5 k 4T 4
q dl 
15c 2 h3
or
qb
(e)
bl
(e)
 T 4
[2.7.11]
[2.7.12]
Where  (Stefan-Boltzmann constant) = 5.676 x 10-8 Wm-2K-4.
2.7.1.6 The solid angle
Consider a hemisphere of radius
r surrounding a differential area
dA1at the center. Fig. 2.7-4a shows
only one-quarter of the hemisphere.
On the hemisphere
dA2  (ab)(cd )
 (r sin  d )(rd )
[2.7.13]
50
The solid angle that intersects dA2 on the hemisphere, shown in Fig.2.7-4b,
is defined as:
dA2
[2.7.14]
d  2  sin dd
r
The energy leaving dA1 in the direction given by
the angle θis IbdA1cosθ, where Ib is the blackbody
intensity. The radiation arriving at some area dA2
at a distance r from A1 would be
dA2
( I b dA1 cos  ) 2
r
Where dA2 is constructed normal to the radius vector.
The dA2/r2 represents the solid angle subtended by the area
dA2 which can be expressed as
dA2  (r sin  d )(rd )
so that
q dA1  
(e)
b
2
0
 I b dA1 
2
0

 /2
0

 /2
0
I b dA1 cos 
r sin  d rd
r2
sin  cos  d d   I b dA1
J. P. Holman, “Heat Transfer”, 1977, p.285
r1
dθ
r1dθ
r2
r2dθ
51
This yields the following equation:
qb( e )   I b
[2.7-18]
and
Ib 
qb( e )

2.7.2 Radiation between blackbodies
Consider two black surface A1 and A2
shown in Fig. 2.7-6. Define the view factors:
F12= fraction of energy leaving surface 1
that reaches surface 2
F21= fraction of energy leaving surface 2
that reaches surface 1
52
As such, the energy leaving surface 1 that reaches surface 2 per unit time is
(e)
Q12  qb1 A1 F12
[2.7-20]
Similarly, the energy leaving surface 2 that reaches surface 1 per unit time is
(e)
Q21  qb 2 A2 F21
[2.7-21]
Therefore, the net energy exchange rate is as follows:
(e)
(e)
Q12  Q12  Q21  qb1 A1 F12  qb 2 A2 F21
[2.7-22]
If both surfaces are at the same temperature T, there is no net energy exchange
and Q12=0. Then substituting Eq.[2.7-12] into Eq. [2.7-22], we have
Q12   T 4  A1F12  A2 F21   0
[2.7-23]
so that
A1F12  A2 F21
[2.7-24]
Which is called the reciprocity relationship. Substituting Eqs. [2.7-12] and [2.7-24]
into [2.7-22], we get
Q12  A1 F12 T14  T2 4   A2 F21 T14  T2 4 
[2.7-25]
53
Let us now proceed to determine the view factors F12 and F21 between the
two black surfaces. From Eq. [2.7-19]
[2.7-26]
qb(1e )
dQ12 
cos 1dA1d 12

where dΩ12 is the solid angle subtended by dA2 as seen from dA1. Let r12 be the
distance between the centers of dA1 and dA2. Consider the hemisphere of radius
r12 and centered at dA1. The projection of dA2 on the hemisphere is cos2dA2.
Therefore, from Eq.[2.7-14]
cos  2 dA2
d 12 
[2.7-27]
r12 2
Substituting Eq.[2.7-27] into Eq. [2.7-26]
dQ12  qb(1e )
And so
Q12  qb(1e ) 
cos 1 cos 2 dA1dA2
 r12 2
cos 1 cos  2
dA1dA2
 r12 2
[2.7-28]
[2.7-29]
Similarly it can be shown that
Q21  qb( e2) 
cos 1 cos  2
dA1dA2
 r12 2
[2.7-30]
54
Substituting Eq. [2.7-29] and Eq. [2.7-30] respectively into Eqs. [2.7-20] and
[2.7-21], we get Eqs. [2.7-31] and [2.7-32]
1 cos 1 cos  2
F12  
dA1dA2
2
A1
 r12
[2.7-31]
and
1
F21 
A2
cos 1 cos  2
  r122 dA1dA2
[2.7-32]
The integration in Eq. [2.7-31] and [2.7-32] is often difficult and needs to be
done numerically. Analytical equations are available for a number of special cases.
Some examples are given below:
55
For large (infinite) parallel plates, long (infinite) concentric cylinders and
concentric spheres F12 = 1, as shown in Fig.2.7-7.
56
For the two identical, parallel directly opposed rectangles shown in Fig. 2.7-8.
  1  X 2 1  Y 2 1 2



2  
X
Y
2
1
2
1
1
1 
  X 1  Y tan
F12 
 Y 1  X tan
 X tan X  Y tan Y 
ln 
2
2
 XY   1  X 2  Y 2 
1

Y
1

X


 

[2.7-33]
where X and Y are defined in the figure.
57
For the two parallel concentric circular disks shown in Fig. 2.7-9,
2



R
1
F12 
X  X 2  4 2  
2
 R1  

[2.7-34]
where X, R1, and R2 are defined in the figure.
For the sphere and the disk shown in Fig. 2.7-10,
1
1 

F12  1 
2
2
1  R2 

[2.7-35]
58
Example
Consider the radiation from the small area dA1 to the flat disk A2, as shown
in the Fig. The element of area dA2 is chosen as the circular ring of radius x.
Calculate FdA1A2.
59
2.7.3 Radiation between graybodies
In radiation heat transfer between blackbodies, all the radiant energy that strikes
a surface is absorbed. In radiation heat transfer between nonblackbodies, the
energy striking a surface will not all be absorbed; part will be reflected. Since the
energy emitted and the energy reflected by a nonblack surface both contribute to
the total energy leaving the surface (J), we can write
J  q ( e )  r q (i )   qb ( e )  r q (i )
[2.7-36]
Where J is called radiosity, is the total energy leaving a surface per unit area per
unit time and r, called the reflectivity, is the fraction of the incident energy reflected.
From the definition for J, we see that
J - q(i) = net energy leaving a surface per unit area per unit time
and
q(i) - J= net energy received at a surface per unit area per unit time
For an opaque material the incident energy is either reflected or absorbed:
r +  =1
[2.7-37]
60
J  q (e)  r q (i )
J q
If Kirchhoff’s law can be applied, that is,  = 
(i )
q

according to Eq. [2.7-8], Eq.[2.7-37] becomes
r
r+=1
[2.7-38]
(e)

J   qb( e )
r
J   qb( e )

1 
From Eqs. [2.7-36] and [2.7-38]
J  q (i )
J   qb ( e ) (1   ) J  J   qb ( e ) qb ( e )  J
J


1 
1 
(1   1)
[2.7-39]
Let us consider radiation heat transfer between two gray surface A1 and A2. Since
the net energy transfer from A1 to A2 (Q12) equals either the net energy leaving A1
or the net energy received at A2, we can write, with the help of Eq. [2.7-39]:
and
1  1
Q12 (
)


(i )
q

J
1


Q12  A1  J1  q1   A1  b1 1   qb(1e )  J1 
 1 1   1 
A1




1 2
(e)
Q
(
)


12
J

q

2
Q12  A2  q2 (i )  J 2   A2  2 b 2   J 2  qb( e2) 
 1  2   1 
A2


(e)
[2.7-40]
[2.7-41]
61
Similar to Eqs. [2.7-20] and [2.7-21] for two black surfaces, we can write the
following expressions for two gray surfaces in terms of the view factors:
Q12  J1 A1F12
[2.7-42]
Q21  J 2 A2 F21
[2.7-43]
(e)
For blackbody Q12  qb1 A1 F12 [2.7-20]
and
Since A1F12 =A2F21 according to Eq. [2.7-24]
Q12  Q12  Q21  A1F12 ( J1  J 2 )
[2.7-44]
From Eqs. [2.7-40], [2.7-44], and [2.7-41]
 1 1   1 
qb1  J1  Q12 

A
1


 1 
J1  J 2  Q12 

A
F
 1 12 
(e)
 1  2   1 
J 2  qb 2  Q12 

A
2


[2.7-45]
[2.7-46]
(e)
[2.7-47]
62
Adding these equations, we have
 1 1   1
1  2  1
1
qb(1e )  qb( e2)  Q12 



A
A
F
A
1
1 12
2


(e)
4
Substituting Eq. [2.7-12] ( qb   T ) into Eq. [2.7-48], we have
 (T  T )  Q12      
4
1
4
2
 (T14  T24 )
Q12
or Q12 
1

[2.7-48]
 (T14  T24 )
[2.7-48a]
For blackbody Q12  A1 F12 T14  T2 4   A2 F21 T14  T2 4 
[2.7-49]
Compared to Eq.[2.7-48a], therefore,
A1F12 
For A1=A2

1

 A2F21 , therefore,
1
1


A1F12 A2F21

1
1
1 1   1 1 1   2 

 


A1F12 A2F21 A1  1
F12
 2 

 1 1 1

1 1
1

1

1


1



1

A  1
 2  A  1  2 
[2.7-50]
63
Example 2.7-3 Radiation shields
Given:
Hot surface T1, 1, cold surface T2, 2, radiation shield T3, 3
Surfaces 1 & 2: two infinite parallel gray surface with the area of A
Find: The radiation heat transfer between two parallel gray surfaces
Since A1=A2=A3, and the surfaces are infinite and parallel
F12=F32=1 (by definition, F13 = fraction of energy leaving
surface 1 and reaches surface 2)
From [2.7-50]
1 1   1
1 3  1


1
1



A F13
A
A1 1
A
1 1
1  1 1 1 

1

1

 1     1

A  1
 3  A  1  3 
1
11 1 
    1
A F32 A   3  2 
Substituting above two equations into Eq.[2.7-49], we get

64
A (T14  T34 )
A (T34  T24 )
Q13 
and Q32 
1 1

1 1



1


1








3
2
 1

 3

Since Q13=Q32=Q132, we have


Q132  1 1
Q132  1 1
4
4
T T 
   1 and T3  T2 
   1
A  1  3 
A   3  2 
4
1
4
3
Adding the two equations, we have
 1 1

Q132  1 1
T T 
   1      1  
A  1  3    3  2  
4
1
4
2
and
Q132 
A T14  T24 
 1 1
 1 1

   1      1  
  1  3    3  2  
65
Without a radiation shield, from Eqs [2.7-49] and [2.7-50]
A (T14  T24 )
Q12 
1 1



1


 1  2 
The ratio of radiation heat transfer with a shield to that without one is
Q132
Q12
If 1=2=3=
Q132
Q12
1 1

   1
 1  2 

1 1
 1 1

   1     1
 1  3    3  2 
2 
  1
1
 
 

2
2
1  1 2


66
Example: Two very large parallel planes with emissivities 0.3 and 0.8 exchange heat.
Find the percentage reduction in heat transfer when a polished-aluminum radiation
shield (=0.04) is placed between them.
Q12  (T14  T24 )

Sol: The heat transfer without the shield is given by A  1 1



1


 1  2 
The radiation network for the problem with the
shield is placed, based on Eq. [2.7-50], is given
E’b1
E b1
E’b3
J1
1-1
1
J3
1
F13
J 3'
3
1- 3
3
J2
1
F32
E b2
1- 2
2
The total resistance is 2.333+2 (24) + 2(1) + 0.25 =52.583
Q12  (T14  T24 )

=0.01712 (T14  T24 )
A
52.583
1 2


1
1
1

 1 
 2
A1F12 A2F21
A1
A1 F12
A2
E’b2
Eb 3
1- 3
1  1
1  1

1  0.3
 0.333
0.3
1
1   3 1  0.04

 24
3
0.04
1   2 1  0.8

 0.25
2
0.8
The heat transfer is reduced by 93.6%
67
Summary
1. Emission and absorption of a black body and a gray body
(e)
qemission
  qb( e )
qb( e )   T14
T1
T1
(black body)
(gray body)
q
(a)
absorption
 q
and
(i )
and
 
2. Radiation between black bodies (Q12: The net energy exchange rate )
(black body)
T2, A2
T1, A1
(black body)
Q12  A1 F12 (T14  T24 )  A2 F21 (T14  T24 )
Q12 Eb1  Eb 2  (T14  T24 )

=
1
1
A
F12
F12
E b1
Eb 2
1
F12
68
Summary
J  q( e)  r q(i )
3. Radiation between gray bodies
(J:total energy leaving a surface/time.cm2 )
(gray body)
J q
T2, A2
(i )
qb( e )  J

1
(  1)

(J  q (i ) :net energy leaving a surface/time.cm2 )
T1, A1
Q12  A1 F12 (T14  T24 )  A2 F21 (T14  T24 )
(gray body)
1  1
E b1
E1
E b2
J1
1-1
1
J2
1
F12
E2

1
1

 1
A1 F12 A2 F21
A1
1 1 2
F

+ 12 + 2
A1
A1
1- 2
2
69